Solution : Let the number, n be 4.

Then, 4²= 16
A.T.Q - n²-n
so, 4²-4 = 16-4 = 12 ( on putting the value of n )

Now to check divisibility by 6 the last digit should be a number divisible by 2 and sum of the numbers should be divisible by 3.

Since , 12 is divisible by both 2 and 3 so it is also divisible by 6
hence proved.

**Proof:**

To prove that n² - n is divisible by 6 for any positive integer n, we need to show that (n² - n) is divisible by both 2 and 3.

**Divisibility by 2:**

Let's consider the two possible cases for n:

**Case 1: n is even**

If n is even, we can write it as n = 2k, where k is a positive integer.

Substituting this into the expression n² - n:

n² - n = (2k)² - (2k) = 4k² - 2k

Factoring out 2 from both terms:

n² - n = 2(2k² - k)

Since 2k² - k is an integer, we can let it be represented by another positive integer m:

n² - n = 2m

Thus, when n is even, n² - n is divisible by 2.

**Case 2: n is odd**

If n is odd, we can write it as n = 2k + 1, where k is a positive integer.

Substituting this into the expression n² - n:

n² - n = (2k + 1)² - (2k + 1) = 4k² + 4k + 1 - 2k - 1

Simplifying:

n² - n = 4k² + 2k = 2(2k² + k)

Similar to the previous case, we can represent 2k² + k as another positive integer m:

n² - n = 2m

Thus, when n is odd, n² - n is also divisible by 2.

Since n² - n is divisible by 2 for both even and odd n, we have established that it is divisible by 2.

**Divisibility by 3:**

Now, let's prove that n² - n is divisible by 3:

We can write n² - n as n(n - 1).

Consider the three possible cases for n:

**Case 1: n is divisible by 3**

If n is divisible by 3, we can write it as n = 3k, where k is a positive integer.

Substituting this into the expression n(n - 1):

n(n - 1) = 3k(3k - 1) = 3(3k² - k)

Since 3k² - k is an integer, we can let it be represented by another positive integer m:

n(n - 1) = 3m

Thus, when n is divisible by 3, n² - n is divisible by 3.

**Case 2: n has a remainder of 1 when divided by 3**

If n has a remainder of 1 when divided by 3, we can write it as n = 3k + 1, where k is a positive integer.

Substituting this into the expression n(n - 1):

n(n - 1) = (3k + 1)(3k + 1 - 1) = (3k + 1)(3k) = 3(3k² + k) + k

Since 3k² + k is an integer, we can let